If you note that a square can be circumscribed by a circle (circumscribed is when the object is drawn in...right?) this makes things handy.
Draw a circle with the compass, for the diameter = side length of the square.
Drawing the circle ensures that the side lengths will all be congruent.
Now take your straightedge, and draw 4 sides, 2 sides parallel to each other on 2 outer sides of the circle, and 2 on the top, also parallel, BUT also perpendicular to the other 2 lines. You have a square
9514 1404 393
Answer:
136
Step-by-step explanation:
The x-value at one end of the beam will be where f(x) = 0.
0 = -x²/272 +17
x² = 17(272) = 4624
x = ±√4624 = ±68
The beam extends from x=-68 to x=68, so has a width of ...
68 -(-68) = 136
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Note that the x in the equation for f(x) is not the same as the x in the picture corresponding to width.
The answer is -420 (negative four-hundred and twenty)